Question

In the given figure, OO is right-angled.

Answer 1

Applying Pythagoras theorem in right-angled $△POR$, we have:

${PR}^2={PO}^2+{OR}^2$

${PR}^2=6^2+8^2=36+64=100$

$\Rightarrow PR=\sqrt{100}=10$cm

In $△PQR,$

${PQ}^2+{PR}^2={24}^2+{10}^2=576+100=676$

And ${QR}^2={26}^2=676$

$\therefore △PQR$ is right-angled at $P$